Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
struct node {
  char ch[10];
  int num;
} opera[400000];
int n, m, a[110], b[110], maze[110][110];
int main() {
  int i, j;
  while (scanf("%d%d", &n, &m) != EOF) {
    int minx, miny, p, k = 1;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        scanf("%d", &maze[i][j]);
      }
    }
    if (n <= m) {
      p = 1;
      for (i = 1; i <= n; i++) {
        minx = inf;
        for (j = 1; j <= m; j++) {
          minx = min(minx, maze[i][j]);
        }
        a[p++] = minx;
      }
      int k = 1;
      for (i = 1; i <= n; i++) {
        if (a[i]) {
          for (j = 1; j <= m; j++) maze[i][j] -= a[i];
          while (a[i]) {
            strcpy(opera[k].ch, "row");
            opera[k++].num = i;
            a[i]--;
          }
        }
      }
      p = 1;
      for (j = 1; j <= m; j++) {
        miny = inf;
        for (i = 1; i <= n; i++) {
          miny = min(miny, maze[i][j]);
        }
        b[p++] = miny;
      }
      for (j = 1; j <= m; j++) {
        if (b[j]) {
          for (i = 1; i <= n; i++) maze[i][j] -= b[j];
          while (b[j]) {
            strcpy(opera[k].ch, "col");
            opera[k++].num = j;
            b[j]--;
          }
        }
      }
      int flag = 0;
      for (i = 1; i <= n; i++) {
        for (j = 1; j <= m; j++)
          if (maze[i][j]) flag = 1;
      }
      if (flag == 1)
        printf("-1\n");
      else {
        printf("%d\n", k - 1);
        for (i = 1; i < k; i++) printf("%s %d\n", opera[i].ch, opera[i].num);
      }
    } else {
      p = 1;
      for (j = 1; j <= m; j++) {
        miny = inf;
        for (i = 1; i <= n; i++) {
          miny = min(miny, maze[i][j]);
        }
        b[p++] = miny;
      }
      for (j = 1; j <= m; j++) {
        if (b[j]) {
          for (i = 1; i <= n; i++) maze[i][j] -= b[j];
          while (b[j]) {
            strcpy(opera[k].ch, "col");
            opera[k++].num = j;
            b[j]--;
          }
        }
      }
      p = 1;
      for (i = 1; i <= n; i++) {
        minx = inf;
        for (j = 1; j <= m; j++) {
          minx = min(minx, maze[i][j]);
        }
        a[p++] = minx;
      }
      for (i = 1; i <= n; i++) {
        if (a[i]) {
          for (j = 1; j <= m; j++) maze[i][j] -= a[i];
          while (a[i]) {
            strcpy(opera[k].ch, "row");
            opera[k++].num = i;
            a[i]--;
          }
        }
      }
      int flag = 0;
      for (i = 1; i <= n; i++) {
        for (j = 1; j <= m; j++)
          if (maze[i][j]) flag = 1;
      }
      if (flag == 1)
        printf("-1\n");
      else {
        printf("%d\n", k - 1);
        for (i = 1; i < k; i++) printf("%s %d\n", opera[i].ch, opera[i].num);
      }
    }
  }
  return 0;
}